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I am currently working on the understanding of the stochastic nature of the Schroedinger equation. This has a notable history dating back to Nelson's works and relative criticisms. But one can take a different path and, starting from a random walk process with probability

\begin{equation} P(k;N) = \binom{N}{k}\left(\frac{1}{2}\right)^N, \end{equation}

one can assume the existence of a "square root" of this process with a complex amplitude

\begin{equation} A(k;N) = \binom{N}{k}^\frac{1}{2}\left(\frac{1}{2}\right)^{N/2}e^{i\phi(k,N)} \end{equation}

such that $P(k;N)=|A(k;N)|^2$ and $\phi(k,N)$ are some phases exactly determined. These are quantum amplitudes. Is this anything making sense? Does any literature exist about?

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    $\begingroup$ In the classical case, $P(k; N)$ is the probability at time $N$ that the walker is at position $k - (N - k)$. I'm assuming that in the quantum case, $A(k; N)$ is meant to be the amplitude at time $N$ for the walker to be at position $k - (N - k)$. This is certainly "something making sense" in that for each time $N$, you've given a quantum state for the walker, so you can calculate the expected value of an observable at any time. $\endgroup$
    – Vectornaut
    Commented Nov 30, 2011 at 20:03
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    $\begingroup$ One question you might ask is whether there's a unitary map that sends the state at time $N$ to the state at time $N + 1$ for all $N$. If there is, you can then ask whether this map is the evolution operator of a quantum random walk that's been studied before. $\endgroup$
    – Vectornaut
    Commented Nov 30, 2011 at 20:04

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Note: This mostly debunks an answer posted by the OP and now deleted. As a consequence of this unfortunate deletion, the argument below might be a little difficult to follow.

As explained on the MSE page you are referring to, very basic arguments show that the limits of the sums $S_n$ you write in your "answer" cannot exist.

Consider for example the second version of $S_n$ (the one based on absolute values) in the simplest possible situation, that is, on the interval $[0,1]$ and when $G\equiv1$. Note that the Gaussian random variables $W(t_i)-W(t_{i-1})$ are independent with variances $t_i-t_{i-1}$. Hence, for every $i$, $$ E((W(t_i)-W(t_{i-1}))^2)=t_i-t_{i-1}, $$ and, for every $i\ne k$, $$ E(|W(t_i)-W(t_{i-1})|\cdot|W(t_k)-W(t_{k-1})|)=c\cdot\sqrt{t_i-t_{i-1}}\cdot\sqrt{t_k-t_{k-1}}, $$ where it happens that $c=2/\pi$ but this is irrelevant. Again in the simplest case, that is, when $t_i=i/n$ for every $i$, this shows that $$ E(S_n^2)=1+c\cdot(n-1). $$ Hence, in contradiction with what you assert, the sequence $(S_n)_{n\geqslant1}$ diverges in $L^2$.

To sum up, integrals such as $$ \int G\cdot|\mathrm dW|\qquad\text{or}\quad\int G\cdot(\mathrm dW)^{\alpha}, $$ whatever their precise definitions would be, cannot exist due to the (ir)regularity properties of the Brownian paths (except, of course, the second one when $\alpha=1$).

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  • $\begingroup$ It is not clear to me how do you cope with the case $(dW)^2$ that is note to have a finite limit. I mean $\int_{t_0}^tG(t')(dW(t'))^2=\int_{t_0}^tG(t')dt'$. This is a particular case of $\int G(dW)^\alpha$ and is well defined in stochastic calculus. We can push this further by considering the case with $(dW)^n$ with $n\in\mathbb{N}$ and $n\ge 3$ that gives 0 for the Ito ontegral. So, your argument is not always true and I am asking when it is not with $\alpha$ real. $\endgroup$
    – Jon
    Commented Jan 26, 2012 at 11:29
  • $\begingroup$ Are you claiming that Ito integral does not exist? $\endgroup$
    – Jon
    Commented Jan 26, 2012 at 11:33
  • $\begingroup$ Now I understand what is wrong with your argument. The limit must be taken in the rms sense $\lim_{n\rightarrow\infty}\rangle(S_n-S)\langle$. Your upvotes should be removed. $\endgroup$
    – Jon
    Commented Jan 26, 2012 at 11:50
  • $\begingroup$ Sorry, I meant $\lim_{n\rightarrow\infty}\langle(S_n-S)^2\rangle$. $\endgroup$
    – Jon
    Commented Jan 26, 2012 at 11:51
  • $\begingroup$ You claim $E(|W(t_i)-W(t_{i-1}||W(t_k)-W(t_{k-1}|)\ne 0$ with $i\ne k$ but increments are not correlated. Could you explain this claim improving your answer, please? $\endgroup$
    – Jon
    Commented Jan 26, 2012 at 18:46

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